Count all unique elements in a 3d matrix

Hi all,

I have created a 3-d matrix randworld and would like to list the unique elements, count the number of unique elements and count the size of the unique elements for randworld(:,:,1), randworld(:,:,2) and randomworld(:,:,3). e.g. 4, 7, 13 is one unique element in this matrix with size 3.

randworld(:,:,1) =

     4     4     3    11    11  
     4     9     3     9    10    
     2     7     9     3     9    
     7     6     6     9     3  
     4     1    15    15    13

randworld(:,:,2) =

     7     7     6     3     3    
     7    11     6     6    10  
    11     4    11     9     6    
     4    15    15     6     9   
     1     8     5     5     5

randworld(:,:,3) =

    13    13     9     4     4   
    13    11     9     3     6    
    13     4    11    12     3     
     4     1     1     3    12   
    11     5    15    15    15

Any help will be appreciated.

Thanks, Vishal

[Azzi Abdelmalek]

why 4,7,13 is unique element, what about other numbers?

[Jan]

I do not understand: "4, 7, 13 is one unique element in this matrix with size 3." I see 9 elements of the value 4.

Could you provide a real example for the outputs for a [3x3x3] array?

[Vishal]

Sorry, I'll try and make it my question more clear. In the 3-d matrix randworld as listed below, I am trying to find the following:

randworld(:,:,1) =

     4     4     3    
     4     9     3         
     2     7     9

randworld(:,:,2) =

     7     7     6     
     7    11     6   
    11     4    11

randworld(:,:,3) =

    13    13     9   
    13    11     9      
    13     4    11

1. list the unique elements-

4,7,13

2,11,13

9,11,11

7,4,4

3,6,9

9,11,11

2. count the number of unique elements

4,7,13 (count is 3) & 2,11,13 (count is 1) & 9,11,11 (count is 1) & 7,4,4 (count is 1) & 3,6,9 (count is 2)& 9,11,11 (count is 1)

Answer 1

[Azzi Abdelmalek]

a1=randworld(:,:,1)
a2=randworld(:,:,2)
a3=randworld(:,:,3)
v=cell2mat(arrayfun(@(x1,x2,x3) [x1 x2 x3],a1(:),a2(:),a3(:),'un',0));
[a,b,c]=unique(v,'rows','stable')
idx=histc(c,(1:size(a,1))')
% a represent unique vectors
% idx represent the repetition of each vector

[Vishal]

works fine. thanks a ton

[Evan]

If that answer solved your problem, be sure to accept it! ;)

[Azzi Abdelmalek]

Or simply

a1=randworld(:,:,1);
a2=randworld(:,:,2);
a3=randworld(:,:,3);
v1=[a1(:) a2(:) a3(:)];
[a,b,c]=unique(v,'rows','stable')
idx=histc(c,(1:size(a,1))')

[Vishal]

Yes, that works fine as well. Just a small typo. In line 4 replace v with v1. Thanks mate.

a1=randworld(:,:,1); a2=randworld(:,:,2); a3=randworld(:,:,3); v1=[a1(:) a2(:) a3(:)]; [a,b,c]=unique(v1,'rows','stable') idx=histc(c,(1:size(a,1))')

Answer 2

[Jan]

It sounds like a basic task for unique and histc. But currently I do not understand the needs exactly.

[Azzi Abdelmalek]

I think, he wants to form vectors from each chanel:

vector1= [a(1,1,1) a(1,1,2) a(1,1,3)]
vector2=[a(1,2,1) a(1,2,2) a(1,2,3)]
and so on

Answer 3

[dpb]

Permutation on above...

rpr=permute(r,[1 3 2]);
r2d=rpr(:,:,1);for i=2:size(rpr,3),r2d=[r2d; rpr(:,:,i)];end
u=unique(r2d,'rows','stable');

[Vishal]

This solutions works fine if the task is to just list the unique rows. Thanks

[dpb]

Well, one presumes one would use the u as in the previous to determine the rest having found them. Didn't see any point in repeating that.